BS EN 62305-4 Electrical and electronic systems within structures Inspection and maintenance of an LPMS The object of the inspection is to verify the following: ● The LPMS complies with the design ● The LPMS is capable of performing its design function ● Additional protection measures are correctly integrated into the complete LPMS Inspections at the implementation stages of an LPMS are particularly important, as LEMP protection measures such as equipotential bonding are no longer accessible after construction has been completed The frequency of the periodical inspections should be determined with consideration to: ● The inspection comprises checking and updating the technical documentation, visual inspections and test measurements The local environment, such as the corrosive nature of soils and corrosive atmospheric conditions ● The type of protection measures employed Visual inspections are very important, and should verify, for example, if bonding conductors and cables shields are intact and appropriate line routeings are maintained Following the inspections, all reported defects should be immediately corrected A visual inspection should also verify that there are no alterations or additions to an installation, which may compromise the effectiveness of the LPMS For example, an electrical contractor may add a power supply line to external CCTV cameras or car park lightning As this line is likely to cross an LPZ, suitable protection measures (eg SPD) should be employed to ensure the integrity of the complete LPMS is not compromised Care should be taken to ensure that SPDs are re-connected to a supply if routine electrical maintenance such as insulation or “flash” testing is performed SPDs need to be disconnected during this type of testing, as they will treat the insulation test voltage applied to the system as a transient overvoltage, thus defeating the object of the test As SPDs fitted to the power installation are often connected in parallel (shunt) with the supply, their disconnection could go unnoticed Such SPDs should have visual status indication to warn of disconnection as well as their condition, which aids the inspection Inspections should be carried out: ● During the installation of the LPMS ● After the installation of the LPMS ● Periodically thereafter ● After any alteration of components relevant to the LPMS ● After a reported lightning strike to the structure Successful management of an LPMS requires controlled technical and inspection documentation The documentation should be continuously updated, particularly to take account of alterations to the structure that may require an extension of the LPMS Summary Damage, degradation or disruption (malfunction) of electrical and electronic systems within a structure is a distinct possibility in the event of a lightning strike Some areas of a structure, such as a screened room, are naturally better protected from lightning than others and it is possible to extend the more protected zones by careful design of the LPS, direct equipotential bonding of metallic services such as water and gas, and equipotential bonding metallic electrical services such as power and telephone lines, through the use of equipotential bonding SPDs An LPS according to BS EN 62305-3 which only employs equipotential bonding SPDs provides no effective protection against failure of sensitive electrical or electronic systems However it is the correct installation of coordinated SPDs that protect equipment from damage as well as ensuring continuity of its operation – critical for eliminating downtime Each of these measures can be used independently or together to form a complete LPMS Careful planning of equipment location and cable routeing also help achieve a complete LPMS For effective protection of electronic equipment and systems, an LPMS requires continual, documented inspections and, where necessary, maintenance in accordance with an LPMS management plan 90 BS EN 62305-4 | Summary www.furse.com Design examples Design examples Introduction 92 Example 1: Country house 94 Example 2: Office block 101 Example 3: Hospital 112 91 www.furse.com Design examples Design examples Design examples Introduction The following section of this guide takes all the aforementioned information and leads the reader through a series of worked examples In Example and the long hand risk management calculations are explained The results determine whether protection measures are required The first example illustrates various possible solutions The next example takes the reader through a complete implementation of the design protection measures 92 It takes the results from the risk calculation and shows how to carry out the requirements of BS EN 62305-3 for the structural aspects and additionally the necessary measures of BS EN 62305-4, for the protection of the electrical and electronic systems housed within the structure Design examples Finally, there is a third example where the evaluation of R4 (economic loss) is reviewed and discussed The first is a simple example of a small country house located in Norfolk, England, and is treated as a single zone R1 – risk of loss of human life is evaluated The next example is an office building near King’s Lynn in Norfolk In this example the structure is split into distinct zones, where the risk components are calculated for each zone By splitting the structure into zones, the designer can pinpoint precisely where (if any) protection measures are required R1 and R2 have been evaluated in this case to ascertain whether there is a risk of loss of human life (R1) as well as illustrating the need for coordinated SPDs as part of the required protection measures (R2) www.furse.com The third example is a hospital situated in the south east of London and again is split into distinct zones R1 and R4 (economic loss) are evaluated the latter of which confirms the cost effectiveness of installing lightning protection measures compared to the potential consequential losses that could be incurred, without any protection Clearly, the majority of times this information will simply not be available to the designer In these events the designer will choose the probability value of one (as given in the appropriate table), which will produce a more conservative solution It will become obvious that this long hand method is both laborious and time consuming, particularly for those people involved in the commercial world of lightning protection With the aid of the software it will be very easy and become routine in nature to automatically calculate the risks R1 and R2 If it is a listed building or has any cultural importance then R3 can additionally be calculated at the same time Furse have therefore developed their own in-house software, which will carry out all the necessary calculations in a fraction of the time and will provide the designer with the optimum solution It will become apparent to everyone who tackles the risk calculations that a lot of detailed information is required for both the structure and the services supplying the structure The more accurate the details are, the more precise will be the recommended protection measures When the designer has completed the risk assessment calculation, the proposed protection measures should be a reflection of the most suitable technical and economic solution BS EN 62305-3 and BS EN 62305-4 then give specific guidance on how to implement these measures Typically, specific details relating to the characteristics of internal wiring (KS3), the screening effectiveness of the structure (KS1) and of shields internal to the structure (KS2) are required to determine probability PMS Whether the internal wiring uses unshielded or shielded cables is another factor that is taken into consideration 93 www.furse.com Design examples Design examples Example 1: Country house Comment Symbol Value – L b , Wb , H b 15, 20, Isolated Cd Rural Ce L = 15m Line environment factor W = 20m LPS None PB Shield at structure boundary None KS1 Shield internal to structure None KS2 People present outside the house None PA Soil resistivity (Ωm) ρ 100 Lightning flash density 1/km /year Ng 0.7 Consider a small country house (see Figure 6.1) near King’s Lynn in Norfolk The structure is situated in flat territory with no neighbouring structures It is fed by an underground power line and overhead telecom line, both of unknown length The dimensions of the structure are: H = 6m In this specific example the risk of loss of human life R1 in the structure should be considered Telephone line (overhead) 15m Parameter Dimensions (m) Location factor Table 6.1: Characteristics of the structure and its environment 20m Parameter Value – Lc 1,000 Height (m) Buried Hc - Transformer None Ct Line shielding None PLD Internal wiring precaution Figure 6.1: Country house Symbol Length (m) LV line (buried) Comment None KS3 Uw = 2.5kV KS4 0.6 None PSPD Withstand of internal system SPD Protection Assigned values The following tables identify the characteristics of the structure, its environment and the lines connected to the structure ● Table 6.1: Characteristics of the structure and its environment ● Table 6.2: Characteristics of incoming LV power line and connected internal equipment ● Table 6.3: Characteristics of incoming telecom line and connected internal equipment The equation numbers or table references shown subsequently in brackets relate to their location in BS EN 62305-2 Table 6.2: Characteristics of incoming LV power line and connected internal equipment Parameter Comment Symbol Value Length (m) – Lc 1,000 Height (m) – Hc Line shielding None PLD Internal wiring precaution None KS3 Uw = 1.5kV KS4 None PSPD Withstand of internal system SPD Protection Table 6.3: Characteristics of incoming telecom line and connected internal equipment 94 Design examples | Example 1: Country house www.furse.com Definition of zones Collection areas The following points have been considered in order to divide the structure into zones: Calculate the collection areas of the structure and the power and telecom lines ● The type of floor surface is different outside to inside the structure a) Collection area of the structure Ad ● The type of floor surface is common within the structure ● The structure is a unique fireproof compartment ● No spatial shields exist within the structure ● Both electrical systems are common throughout the structure A d = Lb × Wb + Hb ( Lb + Wb ) + π (3 Hb )2 A d = 15 × 20 + × 6(15 + 20) + π (3 × 6)2 A d = 300 + 1, 260 + 1, 018 A d = 2, 578 m2 The following zones are defined: ● Z1 (outside the building) ● Z2 (inside the building) (E A.2) b) Collection area of the power line Al(P) If we consider that no people are at risk outside the building, risk R1 for zone Z1 may be disregarded and the risk assessment performed for zone Z2 only Characteristics of zone Z2 are reported in Table 6.4 Parameter A l(P) = ρ ⎡ Lc − 3( H a + Hb ) ⎤ ⎣ ⎦ (Table A.3) As the power line is not connected to a structure at end ‘a’ of the line then Ha = As length of the power line is unknown then assume Lc = 1000m Comment Symbol Value Wood ru x 10-5 A l(P) = ρ ( Lc − Hb ) Ordinary rf x 10-2 A l(P) = 100 (1, 000 − × 6) Special hazard None hz Fire protection None rp Internal power systems Yes Connected to LV power line – Internal telephone systems Yes Connected to telecom line – Loss by touch and step voltages Yes Lt x 10-4 Loss by physical damage Yes Lf Floor surface type Risk of fire A l(P) = 9, 820 m2 c) Table 6.4: Characteristics of Zone Z2 (inside the building) Collection area near the power line Ai(P) A i(P) = 25 Lc ρ A i(P) = 25 × 1, 000 × 100 A i(P) = 250, 000 m2 d) Collection area of the telecom line Al(T) ( A l(T) = ⎡ Lc − H a + Hb ) ⎤ H c ⎣ ⎦ The actual risk is now determined in the following calculation stages based on the assigned values From this point on a subscript letter will be added to several factors relating to lines entering the structure This subscript (P or T) will identify whether the factor relates to the Power or Telecom line (Table A.3) (Table A.3) As Ha = and Hc = 6m above ground then A l(T) = H c ( Lc − Hb ) A l(T) = × 6(1, 000 − × 6) A l(T) = 35, 352m2 95 www.furse.com Example 1: Country house | Design examples Design examples e) Collection area near the telecom line Ai(T) A i(T) = 1, 000 × Lc f) (Table A.3) NDa(P) = N g × A d/a × Cd/a × Ct × 10−6 A i(T) = 1, 000 × 1, 000 A i(T) = 1, 000, 000 m Annual number of events to the structure at end of power line NDa(P) (E A.5) NDa(P) = 0.7 × × 1× 1× 10−6 NDa(P) = Number of dangerous events Calculate the expected annual number of dangerous events (ie number of flashes) a) g) Annual number of events to the structure at end of telecom line NDa(T) Annual number of events to the structure ND ND = N g × A d/b × Cd × 10−6 NDa(T) = N g × Ad/a × Cd/a × Ct × 10−6 (E A.4) (E A.5) NDa(T) = 0.7 × × 1× 1× 10−6 ND = 0.7 × 2, 578 × 1× 10−6 NDa(T) = ND = 0.0018 Expected annual loss of human life b) Annual number of events to the power line NL(P) NL(P) = N g × A l(P) × Cd(P) × Ct(P) × 10−6 (E A.7) NL(P) = 0.7 × 9, 820 × 1× 1× 10−6 Loss Lf defines losses due to physical damage applicable to various classifications of structures (eg hospitals, schools, museums) L t = 1× 10−4 (See Table NC.1 – inside building) NL(P) = 0.0069 c) Loss Lt defines losses due to injuries by step and touch voltages inside or outside buildings L f = (See Table NC.1 – House) Annual number of events near the power line NI(P) NI(P) = N g × A i(P) × Ct(P) × Ce(P) × 10−6 a) Calculate loss related to injury of living beings LA LA = × L t (E A.8) NI(P) = 0.7 × 250, 000 × 1× 1× 10−6 LA = 0.00001× 0.0001 NI(P) = 0.175 LA = 1× 10−9 d) Annual number of events to the telecom line NL(T) NL(T) = N g × A l(T) × Cd(T) × Ct(T) × 10−6 NL(T) = 0.7 × 35, 352 × 1× 1× 10 b) Calculate loss in structure related to physical damage (flashes to structure) LB (E A.7) LB = h Z × r p × r f × L f −6 96 LB = 1× 10−2 Annual number of events near the telecom line NI(T) NI(T) = N g × A i(T) × Ct(T) × Ce(T) × 10−6 (E NC.4) LB = 1× 1× 0.01× NL(T) = 0.0247 e) (E NC.2) (E A.8) c) Calculate loss related to injury of living beings (flashes to service) LU LU = r u × L t NI(T) = 0.7 × 1, 000, 000 × 1× 1× 10−6 LU = 0.00001× 0.0001 NI(T) = 0.7 (E NC.3) LU = 1× 10−9 Design examples | Example 1: Country house www.furse.com d) Calculate risk to the power line resulting in physical damage RV(P) d) Calculate loss in structure related to physical damage (flashes to service) LV LV = h Z × r p × r f × L f RV(P) = ( NL(P) + NDa ) PV × LV (E NC.4) LV = 1× 1× 0.01× RV(P) = (0.0069 + 0) × 1× 1× 10−2 LV = 1× 10−2 (E 26) RV(P) = 6.9 × 10−5 e) Loss of human life R1 The primary consideration in this example is to evaluate the risk of loss of human life R1 Risk R1 is made up from the following elements/coefficients Calculate risk to the telecom line resulting in shock to humans RU(T) RU(T) = ( NL(T) + NDa ) P × LU U R1 = RA + RB + RC * + RM * + RU + RV + RW * + RZ * (E 1) RU(T) = (0.025 + 0) × 1× 1× 10−9 * (E 25) RU(T) = 2.5 × 10−11 say RU(T) = Only for structures with risk of explosion and for hospitals with life saving electrical equipment or other structures when failure of internal systems immediately endangers human life Thus, in this case Calculate risk to the structure resulting in shock to humans RA RA = ND × PA × LA (E 21) RA = 0.0018 × × 1× 10−9 RA = 1.8 × 10−12 (E 26) RV(T) = (0.0247 + 0) × 1× 1× 10−2 RV(T) = 2.47 × 10−4 or 24.7 × 10−5 Thus: R1 = RA + RB + RU(P) + RV(P) + RU(T) + RV(T) say RA = b) Calculate risk to the structure resulting in physical damage RB RB = ND × P × LB B Calculate risk to the telecom line resulting in physical damages RV(T) RV(T) = ( NL(T) + NDa ) PV × LV R1 = RA + RB + RU(P) + RV(P) + RU(T) + RV(T) a) f) R1 = + 1.8 + + 6.9 + + 24.7 R1 = 33.4 × 10−5 (E 22) RB = 0.0018 × 1× 1× 10−2 RB = 1.805 × 10−5 c) This result is now compared with the tolerable risk RT for the loss of human life R1 Thus: R1 = 33.4 × 10−5 > RT = 1× 10−5 Calculate risk to the power line resulting in shock to humans RU(P) ( ) RU(P) = NL(P) + NDa P × LU U RU(P) = (0.0069 + 0) × 1× 1× 10−9 RU(P) = 6.9 × 10−12 (E 25) Therefore protection measures need to be instigated The overall risk R1 may also be expressed in terms of the source of damage Source of damage, page 13 R = RD + RI (E 5) Where: say RU(P) = RD = RA + RB (E 6) RD = + 1.8 RD = 1.8 www.furse.com Example 1: Country house | Design examples 97 Design examples Thus: Thus: RD = 1.8 × 10−5 > RT = 1× 10−5 RB = ND × P × LB B Therefore protection measures against a direct strike to the structure need to be instigated RB = 0.0018 × 0.2 × 1× 10−2 And RB = 3.6 × 10−6 RI = RU(P) + RV(P) + RU(T) + RV(T) (E 22) or 0.36 × 10−5 (E 7) Where: RI = + 6.9 + + 24.7 RI = 31.6 Thus: Similarly we need to apply SPDs at the entrance point of the building for the power and telecom lines corresponding with the structural protection measure ie SPDs Type III-IV We therefore assign PV = 0.03 Thus: RV(P) = ( NL(P) + NDa ) PV × LV (E 26) RI = 31.6 × 10−5 > RT = 1× 10−5 RV(P) = (0.0069 + 0) × 0.03 × 1× 10−2 Therefore protection measures against an indirect strike to the structure need to be instigated RV(P) = 2.07 × 10−6 Analysing the component results that make up R1 we can see that RV(T) is by far the largest contributor to the actual risk R1 Similarly: Component RV(T) = 24.7 and R1 = 33.4 or 0.207 × 10−5 RV(T) = ( NL(T) + NDa ) PV × LV (E 26) Thus component RV(T) represents: RV(T) = (0.0247 + 0) × 0.03 × 1× 10−2 ⎛ 24.7 ⎞ ⎜ 33.4 × 100% ⎟ = 73.9% of R1 ⎝ ⎠ RV(T) = 7.41× 10−6 Component RV(P) is next significant contributor to R1 or 0.741× 10−5 Risk Component RV(P) represents: Value x 10-5 RA Solution A To reduce RD we should apply a structural Lightning Protection System and so reduce PB from to a lower value depending on the Class of LPS (I to IV) that we choose By the introduction of a structural Lightning Protection System, we automatically need to install service entrance lightning current SPDs at the entry points of the incoming telecom and power lines, corresponding to the structural Class LPS 98 0.207 RU(T) To reduce the risk to the tolerable value the following protection measures could be adopted: RV(P) RV(T) 0.741 Total Protection measures 0.36 RU(P) RV(T) and RV(P) represent 94.6% of reason why R1 > RT RB ⎛ 6.9 ⎞ ⎜ 33.4 × 100% ⎟ = 20.7% of R1 ⎝ ⎠ 1.308 Risks Ͼ 1x10-5 are shown in red Risks р 1x10-5 are shown in green Table 6.5: Summary of individual risks after first attempt at protection solution A Thus: R1 = 1.308 × 10−5 > RT = 1× 10−5 Therefore additional protection measures need to be instigated This reduces RV(T) and RV(P) to a lower value, depending on the choice of Class of LPS If we apply a structural LPS Class IV, we can assign PB = 0.2 Design examples | Example 1: Country house www.furse.com If we use SPDs with superior protection measures (ie lower let through voltage) for both the telecom and power lines we can apply SPDs of Type III-IV*, ie we can assign PV = 0.003 (see Table NB.3) Thus: RV(P) = ( NL(P) + NDa ) PV × LV (E 26) RV(P) = (0.0069 + 0) × 0.003 × 1× 10−2 RV(P) = 2.07 × 10 −7 or 0.021× 10 Thus: RB = ND × P × LB B (E 22) RB = 0.0018 × 0.05 × 1× 10−2 RB = 9.02 × 10−7 or 0.09 × 10−5 We now need to apply SPDs of Type II at the entrance point of the building for the power and telecom lines, to correspond with the structural protection measure We therefore assign PV = 0.02 −5 Thus: Similarly: RV(T) = ( NL(T) + NDa ) PV × LV (E 26) RV(P) = ( NL(P) + NDa ) PV × LV (E 26) RV(T) = (0.0247 + 0) × 0.003 × 1× 10−2 RV(P) = (0.0069 + 0) × 0.02 × 1× 10−2 RV(T) = 7.41× 10−7 RV(P) = 1.375 × 10−6 or 0.074 × 10−5 or 0.138 × 10−5 Similarly: Risk Value x 10-5 RA RB 0.36 RU(P) RV(P) 0.021 RU(T) RV(T) 0.074 Total 0.455 RV(T) = ( NL(T) + NDa ) PV × LV (E 26) RV(T) = (0.0247 + 0) × 0.02 × 1× 10−2 RV(T) = 4.95 × 10−6 or 0.495 × 10−5 Risk Value x 10-5 RA < RT = 1× 10 Therefore protection has been achieved RV(T) 0.495 Total R1 = 0.455 × 10 −5 0.138 RU(T) −5 RV(P) Thus: 0.09 RU(P) Table 6.6: Summary of individual risks after second attempt at protection solution A RB Risks Ͼ 1x10-5 are shown in red Risks р 1x10-5 are shown in green 0.723 Risks Ͼ 1x10-5 are shown in red Risks р 1x10-5 are shown in green Solution: Table 6.7: Summary of individual risks for protection solution B Install a structural LPS Class IV along with service entrance SPDs of Type III-IV* on both the incoming power and telecom lines Thus: Solution B R1 = 0.723 × 10−5 < RT = 1× 10−5 An alternative approach would be to fit a higher Class of LPS If we now apply a structural LPS Class II, we can assign PB = 0.05 Therefore protection has been achieved Solution: Install a structural LPS Class II along with service entrance SPDs of Type II on both the incoming power and telecom lines 99 www.furse.com Example 1: Country house | Design examples Design examples Solution C If we maintain service entrance SPDs with the lower let through voltage ie SPDs of Type III-IV* on both the incoming telecom and power lines, but this time install manual fire extinguishers, strategically placed throughout the house then rp can be reduced from no fire provision rp = to rp = 0.5 No structural protection is installed Thus: LB = LV = h Z × r p × r f × L f (E NC.4) Thus: R1 = 0.947 × 10−5 < RT = 1× 10−5 Therefore protection has been achieved Solution: Install manual fire extinguishers strategically placed throughout the house and install structural service entrance overvoltage SPDs of Type III-IV* on both the incoming power and telecom lines LB = LV = 1× 0.5 × 0.01× LB = LV = × 10−3 Decision As can be seen by this example of the Country house there are several “protection measure” solutions So: RB = ND × P × LB B (E 22) RB = 0.0018 × 1× × 10−3 RB = × 10 −6 or 0.9 × 10 Another solution is a structural LPS Class II combined with service entrance lightning current SPDs of Type II on both incoming service lines −5 Similarly: RV(P) = ( NL(P) + NDa ) PV × LV (E 26) RV(P) = (0.0069 + 0) × 0.003 × × 10−3 RV(P) = 1.035 × 10−7 or 0.01× 10−5 It is however, the third option of manual fire extinguishers and overvoltage SPDs that is, in this case, the most economic solution RV(T) = ( NL(T) + NDa ) PV × LV (E 26) RV(T) = (0.0247 + 0) × 0.003 × × 10 −3 Value x 10-5 RA RB 0.9 RU(P) RV(P) 0.01 RU(T) RV(T) 0.037 Total 100 Risks Ͼ 1x10-5 are shown in red Risks р 0.947 1x10-5 SPD Recommendations Solution C was deemed to be the most cost effective option, and this involved the installation of fire extinguishers throughout the house, and service entrance overvoltage SPDs on the incoming power and telecom lines or 0.037 × 10−5 Risk A third option is the installation of manual fire extinguishers strategically placed throughout the house and the installation of service entrance overvoltage SPDs of Type III-IV* (ie with a lower let through voltage) on both incoming service lines All three solutions ensure that the actual risk R1 is lower than the tolerable value RT Similarly: RV(T) = 3.705 × 10−7 One option is a structural LPS Class IV combined with service entrance lightning current SPDs of Type III-IV* (ie with a lower let through voltage) on both incoming service lines are shown in green Table 6.8: Summary of individual risks for protection solution C Design examples | Example 1: Country house As no structural LPS is required, and the power cable enters the structure from an underground duct, there is only a need to fit an overvoltage SPD of Type III/IV* The enhanced or* category of SPD indicates that an SPD with a voltage protection level of no more than 600V should be used (see Table 3.5 Note 3, 3rd paragraph on page 30) The power supply is single phase, so an ESP 240 M1 should be installed at the consumer unit, on the load side of the main isolator, housed within a WBX enclosure The telecom cable feeds a single BT socket The incoming telecom cable is overhead and therefore may see partial lightning currents An ESP TN/JP fitted at the BT socket would offer the required level of protection www.furse.com Example 2: Office block Consider a small five storey office block housing an insurance company (see Figure 6.2) near King’s Lynn in Norfolk The structure is situated in flat territory with no neighbouring structures It is fed by an underground power line 650m long and underground telecom line of unknown length The dimensions of the structure are: Parameter Comment Dimensions (m) Value L b , Wb , H b 40, 20, 15 Isolated Cd 0.5 Line environment factor Rural Ce 0.1 LPS None PB Shield at structure boundary None KS1 H = 15m In this specific example the risk of loss of human life R1 and loss of service to the public R2 should be considered Shield internal to structure None KS2 People present inside/outside the structure Yes nt 200 Soil resistivity (Ωm) ρ 250 Lightning flash density 1/km2/year Ng 0.7 L = 40m W = 20m LV line (buried) Location factor Symbol – Table 6.9: Characteristics of the structure and its environment Telephone line (buried) Parameter Comment Symbol Value – Lc 650 Buried Hc - Ct None PLD None KS3 Uw = 2.5kV KS4 0.6 None PSPD Length (m) Height (m) Transformer Internal wiring precaution 20m 40m Figure 6.2: Office block Assigned values The following tables identify the characteristics of the structure, its environment and the lines connected to the structure ● ● ● None Line shielding Withstand of internal system SPD Protection Table 6.10: Characteristics of incoming LV power line and connected internal equipment Parameter Comment Symbol Value Length (m) – Lc 1,000 Height (m) Buried Hc – Table 6.10: Characteristics of incoming LV power line and connected internal equipment Line shielding None PLD Internal wiring precaution None KS3 Table 6.11: Characteristics of incoming telecom line and connected internal equipment Withstand of internal system Uw = 1.5kV KS4 None PSPD Table 6.9: Characteristics of the structure and its environment The equation numbers or table references shown subsequently in brackets relate to their location in BS EN 62305-2 SPD Protection Table 6.11: Characteristics of incoming telecom line and connected internal equipment Throughout this example a numerical subscript will be added to several factors This subscript will identify the major risk to which the factors relate For example loss Lt relating to the risk R2 will be written as Lt2 www.furse.com 101 Example 2: Office block | Design examples Design examples Definition of zones The following characteristics of the structure have been considered in order to divide it into zones: ● The type of floor surface is different in the entrance area, in the garden and inside the structure Parameter Comment Symbol Value Floor surface type Linoleum ru x 10-5 Risk of fire High rf 0.5 Special hazard Low panic hz ● The structure is a unique fireproof compartment Fire protection ● Automatic rp 0.2 The archive within the structure is a unique fireproof compartment Spatial shield None KS2 ● No spatial shields exist within the structure Internal power systems Yes Connected to LV power line – ● Both electrical systems are common throughout the structure Internal telephone systems Yes Connected to telecom line – Loss by touch and step voltages Yes Lt See Expected amount of loss, pages 103-104 Loss by physical damage Yes Lf See Expected amount of loss, pages 103-104 – np – 20 persons hour/day days a week The following zones are defined: ● Z1 – Entrance area to the building see Table 6.12 ● Z2 – Garden see Table 6.13 ● Z3 – Archive see Table 6.14 ● Z4 – Offices see Table 6.15 ● Z5 – Computer centre see Table 6.16 Parameter Comment People potentially in danger in the zone Table 6.14: Characteristics of Zone Z3 (Archive) Symbol Value Soil surface type Marble x 10-3 Shock protection None PA Lt x 10-4 Loss by touch and step voltages People potentially in danger in the zone Comment Symbol Value Floor surface type Linoleum ru x 10-5 rf 0.01 Low panic hz Fire protection Manual rp 0.5 Spatial shield Yes Ordinary Special hazard Risk of fire np Parameter Comment Symbol Grass x 10-2 Shock protection Fence PA Loss by touch and step voltages Yes Lt x 10-4 People potentially in danger in the zone – np Internal power systems Yes Connected to LV power line – Yes Connected to telecom line – Loss by touch and step voltages Yes Lt See Expected amount of loss, pages 103-104 Loss by physical damage Yes Lf See Expected amount of loss, pages 103-104 – np – 160 persons hour/day days a week Value Soil surface type None KS2 Internal telephone systems – Table 6.12: Characteristics of Zone Z1 (Entrance area) 102 Parameter People potentially in danger in the zone Table 6.15: Characteristics of Zone Z4 (Offices) Table 6.13: Characteristics of Zone Z2 (Garden) Design examples | Example 2: Office block www.furse.com ... section of this guide takes all the aforementioned information and leads the reader through a series of worked examples In Example and the long hand risk management calculations are explained The results... added to several factors relating to lines entering the structure This subscript (P or T) will identify whether the factor relates to the Power or Telecom line (Table A. 3) (Table A. 3) As Ha = and... 10−5 An alternative approach would be to fit a higher Class of LPS If we now apply a structural LPS Class II, we can assign PB = 0.05 Therefore protection has been achieved Solution: Install a structural